$ F = \left[\begin{array}{rrr}-2 & 4 & 0 \\ 0 & 5 & 5\end{array}\right]$ $ C = \left[\begin{array}{rr}2 & -1 \\ 4 & 5 \\ 0 & 1\end{array}\right]$ What is $ F C$ ?
Solution: Because $ F$ has dimensions $(2\times3)$ and $ C$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ F C = \left[\begin{array}{rrr}{-2} & {4} & {0} \\ {0} & {5} & {5}\end{array}\right] \left[\begin{array}{rr}{2} & \color{#DF0030}{-1} \\ {4} & \color{#DF0030}{5} \\ {0} & \color{#DF0030}{1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ F$ , with the corresponding elements in column $j$ of the second matrix, $ C$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ F$ with the first element in ${\text{column }1}$ of $ C$ , then multiply the second element in ${\text{row }1}$ of $ F$ with the second element in ${\text{column }1}$ of $ C$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-2}\cdot{2}+{4}\cdot{4}+{0}\cdot{0} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ F$ with the corresponding elements in ${\text{column }1}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{2}+{4}\cdot{4}+{0}\cdot{0} & ? \\ {0}\cdot{2}+{5}\cdot{4}+{5}\cdot{0} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ F$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{2}+{4}\cdot{4}+{0}\cdot{0} & {-2}\cdot\color{#DF0030}{-1}+{4}\cdot\color{#DF0030}{5}+{0}\cdot\color{#DF0030}{1} \\ {0}\cdot{2}+{5}\cdot{4}+{5}\cdot{0} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-2}\cdot{2}+{4}\cdot{4}+{0}\cdot{0} & {-2}\cdot\color{#DF0030}{-1}+{4}\cdot\color{#DF0030}{5}+{0}\cdot\color{#DF0030}{1} \\ {0}\cdot{2}+{5}\cdot{4}+{5}\cdot{0} & {0}\cdot\color{#DF0030}{-1}+{5}\cdot\color{#DF0030}{5}+{5}\cdot\color{#DF0030}{1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}12 & 22 \\ 20 & 30\end{array}\right] $